How to find a fundamental solution of $L(f)=f''-c^2f$ I have been given the following definition of the fundamental solution

Let $L$ be a differential operator with $C^{\infty}$ coefficients.
Then the solution to the equation $L(U_{\xi})=\delta_{\xi}$ is called
a fundamental solution.

I wanted to try this on some example. I would like to find the fundamental solution of
$$L(u) = u''-c^{2} u \quad c \geq 0$$
In my lecture notes it is writen that
If one searches fundamental solutions, which are represented through functions, then these must satisfy
$$\int_{\mathbb{R}} U(x,\xi)L^*(\phi) dx=\phi(\xi) \quad \forall \phi \in \mathcal{D}(\mathbb{R})$$
My working
First I calculated the adjoint differential operator and found that
$\langle L(u),\psi \rangle= \langle u,L(\psi) \rangle$, thus $L$ is self adjoint.
How I understood it, I need to solve
$$L(U(x,\xi))=\delta_{\xi}$$
which means
$$\frac{d^2}{dx^2} U(x,\xi)- c^2 U(x,\xi)=\delta_{\xi}$$
Let $H(x,\xi)$ denote the Heaviside Function, where the 'jump' exists at $\xi \in \mathbb{R}$. I know that $\frac{d}{dx}H(x,\xi)=\delta_{\xi}$, though I am not really sure how I can use this to solve the equation (or if I can use this). I don't know how to tackle this equation. I would be thankful if someone could help me.
Edit
I tried the following
$$L(U(x,\xi))=\delta_{\xi}$$
1.
$$U''(x,\xi)-c^2U(x,\xi)=\delta_{\xi}$$
I think the solution should be split in 3 parts, $x<\xi, x=\xi$ and $x>\xi$. If I solve the equation (1) for $x \neq \xi$, I get something like
$$U(x,\xi)=
\begin{array}{cc}
  \{ & 
    \begin{array}{cc}
      c_1e^{c(x-\xi)}+c_2 e^{-c(x-\xi)} & x < \xi \\
      c_3e^{c(x-\xi)}+c_4 e^{-c(x-\xi)}   & x > \xi \\
    \end{array}
\end{array}
$$
Further I know that if $U(x,\xi )$ is a fundamental solution it must satisfy:
Let $\phi$ be a test function, then
2.
$$\int_{-\infty}^{\infty}U(x,\xi)(\phi''(x)-c^2 \phi(x))dx=\phi(x)$$
I assume there will be a 'jump' at $x=\xi$ but I am not really sure how to get there. Also I am not really sure how to calculate the coefficients $c_1,c_2,c_3,c_4$
I tried to split the integral (2) up in two parts. For $\epsilon>0$
3.
$$\int_{-\infty}^{\xi-\epsilon }U(x,\xi)(\phi''(x)-c^2 \phi(x))dx+\int_{\xi+\epsilon}^{\infty}U(x,\xi)(\phi''(x)-c^2 \phi(x))dx$$
I tried to integrate by parts such that I get something like
$$(\text{something})+\int{\phi'(x)(\text{something else})}dx \tag a$$
for both integrals in (3), in the hope that, since (a) must equal $\phi(x)$, I could use this somehow to get the solution.
My actual result of integration by parts of (3) is:
\begin{align}
&c_1e^{-c \epsilon}+c_2e^{c \epsilon} \phi(\xi - \epsilon)+(-c_1 c e^{-c \epsilon}+c_2 c e^{c \epsilon})\phi(\xi - \epsilon)\\
&+(-c_3e^{\epsilon c}-c_4^{-c \epsilon}) \phi(\xi+\epsilon)+(c_3 c e^{\epsilon c}-c_4 ce^{-\epsilon c})\phi(\xi + \epsilon)\\
&+\int_{\infty}^{\xi- \epsilon}\phi'(x)(c_1ce^{c(x- \xi)}(\xi-x+1)+c_2c^{-c \ (x - \xi)}(x- \xi-1))dx\\
&+\int_{\xi+\epsilon}^{\infty}\phi'(x)(c_3ce^{c(x-\xi)}(\xi - x +1)+c_4c^{e^{-c(x-\xi)}}(x-\xi-1))
\end{align}
 A: Yes, your edit is one way to approach $u''-c^2u=\delta$ (since constant coefficient differential operators are translation invariant, without loss of generality we can consider $\delta$ at $0$).
That is, quite rigorously, away_from$0$_, $\delta$ is locally $0$. The idea that distributions are "local", and do/can have pointwise values (etc.) away from "singular support" is standard, perhaps unglamorous, but does rigorize the otherwise possibly-too-cavalier talk about values of $\delta$. Note that we cannot talk about any "value" of $\delta$ at $0$, because that point is in its singular support...
So, yes, as you realized, you want to patch together a solution to the left, and a solution to the right, to get the $\delta$. As in general with linear equations, there is great ambiguity in the constants, since we can always adjust a solution to the inhomogeneous equation by solutions to the homogeneous one.
(An analogous problem of construction of Green's functions for Sturm-Liouville problems on finite intervals uses the additional boundary-value conditions to eliminate that ambiguity.)
In your example, and with $c>0$, we could attempt to use the best-behaved solutions in $x<0$ and $x>0$, namely, the exponential functions that do not blow up as $x\to -\infty$ and $x\to+\infty$: $e^{cx}$ and $e^{-cx}$, respectively. The symmetry of the problem under $x\to -x$ implies that we can take a linear combination $C\cdot e^{c|x|}$ for some constant.
Indeed, to avoid having the first derivative already have a jump, introducing a $\delta$, we want the limit at $0$ from the left to match that from the right. The jump in first derivatives will produce the $\delta$.
Another way to think about this is to write $D$ for $d/dx$, and the equation as $(D-c)(D+c)u=\delta$. Thus, we can first solve $(D-c)v=\delta$, and then solve $(D+c)u=v$. The equation $(D-c)v=\delta$ can be simplified by letting $v=e^{cx}w$, turning it into $e^{cx}w'=\delta$. We can multiply through by $e^{-cx}$ (compactly supported distributions can rigorously be multiplied by smooth functions), to get $w'=\delta$. So $w$ could be the Heaviside function $H$, and $v=e^{cx}H$. And so on.
